Narayana Solutions iit jee 2010

8/9/2019 Narayana Solutions iit jee 2010

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NARAYANA IIT ACADEMY 3

NARAYANA IIT ACADEMY - INDIAIIT JEE (2010) PAPER I QUESTION & SOLUTIONS ( CODE 0)

PART I : CHEMISTRYPAPER I

SECTI ON ISingle Correct Choice Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. In the reaction OCH 3HBr the products are

(A) OCH 3Br and H2

(B) Br and CH 3Br

(C) Br and CH 3OH

(D) OH and CH 3Br

Key: (D)Sol.: HBr H Br+ +

O CH 3 H O CH 3

H2

BrSN

OH

CH3Br

2. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot thatfollowing Arrhenius equation is

(A)

k

T

(B)

k

T

(C)k

T

(D)k

T Key: (A)

Sol.: aE

RTK Ae=

Rate constant K increases exponentially with the rise in temperature. Since rate const. K also depends uponorientation factor A hence its maximum value is not at all infinity rather limited to an optimal value.

3. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is(A) Br 2 (g) (B) Cl 2 (g)(C) H 2O (g) (D) CH 4(g)


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IIT-JEE2010-Code-0-Quest i ons an d Solut ions-Paper-I and II


Key: (B)Sol.: Bromine and water exist in liquid state at 298 K. Methane is not an elemental species.

4. The ionization isomer of ( ) ( )2 24Cr H O Cl NO Cl is(A) ( ) ( )2 2 24Cr H O O N Cl (B) ( )2 2 24[Cr H O Cl ](NO ) (C) ( ) ( )2

4Cr H O Cl ONO Cl

(D) ( )2 2 2 24

Cr H O Cl ( NO ) H O

Key: (B)

Sol.: ( ) ( ) ( ) ( )ionization2 2 2 24 4Cr H O Cl NO Cl Cr H O Cl NO Cl+

+

( ) ( ) ( )Ionization2 2 2 2 2 24 4Cr H O Cl NO Cr H O Cl NO+

+ .

5. The correct structure of ethylenediaminetetraacetic acid (EDTA) is

(A)

CH2COOH

N CH CH

CH 2COOH

NCH 2

CH 2

COOH

COOH

(B)

HOOC

N CH 2 CH 2HOOC

NCOOH

COOH

(C)

CH2COOH

N CH 2 CH 2

CH 2COOH

NCH 2

CH 2

COOH

COOH

(D)CH2COOH

N CH CH

H

NH

CH2 COOH

H2CCOOH

CH2COOH

Key: (C)Sol.: Based on facts

6. The bond energy (in kcal mol 1) of a CC single bond is approximately(A) 1 (B) 10(C) 100 (D) 1000.

Key: (C)Sol.: C C single bond dissociation energy ranges between 88 to 150 K cal mol 1.

7. The synthesis of 3 octyne is achieved by adding a bromolkane into a mixture of sodium amide and analkyne. The bromoalkane and alkyne respectively are(A)

2 2 2 2 3BrCH CH CH CH CH and

3 2CH CH C CH (B)

2 2 3BrCH CH CH and

3 2 2CH CH CH C CH

(C) 2 2 2 2 3BrCH CH CH CH CH and 3CH C CH (D) 2 2 2 3BrCH CH CH CH and 3 2CH CH C CH .Key: (D)Sol.: CH 3 CH 2 C C C CH 2 CH 2 CH 3

3 octyne

3 2 2 22

CH CH CH CH BrNaNH

3 2 3 2CH CH C CH CH CH C C

3 2 2 2 2 3CH CH CH CH C C CH CH

3 octyne

8. The correct statement about the following disaccharide is

O

OH

H

OCH 2

H

OH

H

H

CH 2OH

OH

H

O

OHH2CO

HOH 2C

CH 2OH

H

H

OH

H

(a) (b)


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(A) Ring (a) is pyranose with glycosidic link (B) Ring (a) is furanose with glycosidic link (C) Ring (b) is furanose with glycosidic link (D) Ring (b) is pyranose with glycosidic link.

Key: (A)Sol.: Ring (a) is pyranose whereas ring(b) is furanose. anomeric form of ring (a) is attached through glycosidic

bond.

SECTION IIMultiple Correct Choice Type

This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9. In the Newman projection for 2, 2 dimethylbutane

Y

CH3 CH 3

X

H H

X and Y can respectively be

(A) H and H (B) H and C 2H5 (C) C 2H5 and H (D) CH 3 and CH 3.Key: (B, D)

Sol.: CH3 C

CH3

CH3

CH2 CH32

2, 2-dimethyl butane

1 3 4

C1 C 2 rotationH

H H

HH H

CH 3 CH 3

C2H5

CH 3CH 3

C2H5

180rotation

X and Y become H and C 2H5

CH3 C

CH3

CH3

CH2 CH3

H H

CH3

CH 3CH 3

CH3

180rotation

CH 3

H H

H3C CH 3

CH 3 X and Y become CH 3 and CH 3.

10. The regent(s) used for softening the temporary hardness of water is (are)(A) 3 4 2Ca (PO ) (B) ( )2Ca OH (C) 2 3Na CO (D) NaOCl .


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Key: (B, C)Sol.: Temporary hardness is due to the presence of bicarbonates of Ca and Mg. Temporary hardness can be

removed by clarkes process which involves the addition of slaked lime( ) ( )3 3 22 2Ca HCO Ca OH 2CaCO 2H O+ +

Washing soda removes both the temporary and permanent hardness.( )3 2 3 3 32Ca HCO Na CO CaCO 2NaHCO+ + .

11. Among the following, the intensive property is (properties are)(A) molar conductivity (B) electromotive force(C) resistance (D) heat capacity.

Key: (A)Sol.: EMF, resistance and heat capacity are extensive properties ofcourse, resistivity is an intensive property.

12. Aqueous solutions of HNO 3, KOH, CH 3COOH, and CH 3COONa of identical concentrations are provided.The pair(s) of solutions which form a buffer upon mixing is(are)(A) 3HNO and 3CH COOH (B) KOH and 3CH COONa

(C) 3HNO and 3CH COONa (D) 3CH COOH and 3CH COONa .

Key: (C, D)Sol.: Mixture of weak acid and its salt are known as acidic buffer.

3 3 2 3 2 3strong acid weak acid

H NO CH CO Na CH CO H Na NO+ + + + +

In an acid based reaction. The equilibrium shifts to the direction which results in the formation of weakeracid.

13. In the reaction

OH

( ) 2Na OH aq / Br the intermediate(s) is (are)

(A)

O

Br

Br

(B)

O

Br Br

(C)

O

Br

(D)

O

Br

Key: (A, C)

Sol.:

OH

is strongly activating towards EAS reaction and it is ortho para directing


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O

Br

Br

O

H Br

O

Br

SECTI ON II ILinked Comprehension Type

This section contains 2 paragraphs. Based upon the first paragraph, 3 multiple choice questions and based upon the second paragraph 2 Multiple choice questions have to be answered. Each of these questions have

four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 14 to 16 Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO 4. 5H 2O), atacamite (Cu 2Cl(OH) 3), cuprite (Cu 2O), copper glance (Cu 2S) andmalachite (Cu 2(OH) 2CO 3). However 80% of the world copper production comes from the ore chalcopyrite (CuFeS 2).The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self reduction.

14. Partial roasting of chalcopyrite produces(A) Cu 2S and FeO (B) Cu 2O and FeO(C) CuS and Fe 2O3 (D) Cu 2O and Fe 2O3

Key: (B)Sol: CuFeS 2 + O 2 Cu 2S + 2FeS + SO 2

The sulphites of copper and iron are partially oxidized

2 22FeS 3O 2FeO 2SO+ +

2 2 2 22Cu S 3O 2Cu O 2SO+ + .

15. Iron is removed from chalcopyrite as(A) FeO (B) FeS(C) Fe 2O3 (D) FeSiO 3

Key: (D)

Sol: Fe is removed in the form of FeSiO 3.2 3FeO SiO FeSiO+

16. In self reduction, the reducing species is(A) S (B) 2O (C) 2S (D) SO 2

Key: (C)Sol: 2 2 2Cu S 2Cu O 6Cu SO+ +

2S oxidized into 4S+ hence it is reducing species .

Paragraph for Questions 17 to 18 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The

resulting potential difference across the cell is important in several processes such as transmission of nerve impulsesand maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

( ) ( ) ( ) ( )M s | M aq; 0.05 molar || M aq; 1molar | M s+ + For the above electrolytic cell the magnitude of the cell potential cellE 70= mV.


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17. For the above cell(A) cellE 0; G 0< > (B) cellE 0; G 0> < (C) cellE 0; G 0< > (D) cellE 0; G 0> <

Key: (B)

Sol: Ecell =2.303RT 0.05

logF 1

= a positive value

= 70 mV (given)Hence G < 0.18. If the 0.05 molar solution of M + is replaced by a 0.0025 molar M + solution, then the magnitude of the cell

potential would be(A) 35 mV (B) 70 mV(C) 140 mV (D) 700 mV

Key: (C)

Sol: cell2.303RT 0.0025

E logF 1

=

= ( )22.303RT log 0.05F

= 2 70 mV= 140 mV.

SECTION IVInteger Answer Type

This Section contains TEN questions. The answer to each question is a Single Digit Integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

19. The value of n in the molecular formula Be nAl2Si6O18 isKey: (3)

Sol.: [ ]126 18Si O

3 2 6 18Be Al Bi O

[ ]126 18Si O

-

-

-

-

-

-

20. The total number of basic groups in the following form of lysine is

H3N CH 2 CH 2 CH 2 CH 2

CH

NH2

C

O

O Key: (2) Sol.: H3N CH 2 CH 2 CH 2 CH 2

CH

NH2 CO

O

* Group are basic.


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Sol.: 235 142 90 192 54 38 0U Xe Sr 3 n + + .

26. In the scheme given below, the total number of intramolecular aldol condensation products formed fromY is

( )32

1 .NaOH aq1. O2. Zn , H O 2. heatY

Key: (1)

Sol.: 32

1. O2. Z n / H O

O

O

( )NaOH aq

OH

O

heat

O

27. The concentration of R in the reaction R P was measured as a function of time and the following data isobtained :

[R] (moloar) 1.0 0.75 0.40 0.10t (min.) 0.0 0.05 0.12 0.18

The order of the reaction isKey: (0) Sol.: R P

dcdt

after 0.05 min = 0.25 50.05

= M min 1

dcdt

after 0.12 min = 060 50.12

= M min 1

dcdt

after 0.18 min = 90 50.18

= M min 1

The average rate remains same throughout. This implies that rate is independent of concentration (zeroorder).

28. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C 4H6 is

Key (5) Sol.: Cyclic isomers C 4H6

CH3

CH3

CH2

Total isomers = 5


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PART II : MATHEMATI CS



29. The number of 3 3 matrices A whose are either 0 or 1 and for which the system Ax 1y 0

z 0

=

has exactly

two distinct solutions, is(A) 0 (B) 2 9 1(C) 168 (D) 2

Key. (A)Sol. Three planes cannot meet only at two dist inct points. Hence A is correct.

30. The value of x

3 4x 00

1 t ln (1 t)lim

x t 4+

+ dt is

(A) 0 (B)1

12

(C)1

24(D)

164

Key. (B)

Sol.

x

40

3x 0

t ln (1 t) dtt 4

limx

++

4 2x 0

x ln (1 x)lim

(x 4) 3x+=

+= 4x 0

ln (1 x)lim

3x(x 4)++

=x 0

1 ln (1 x)lim

4 3 x+

=

112

.

31. Let p and q be real numbers such that p 0, p 3 q. If and are nonzero complex numbers satisfying

+ = p and 3 + 3 = q, then a quadratic equation having

and

as its roots is

(A) (p 3 + q) x 2 (p 3 + 2q) x + (p 3 + q) = 0 (B) (p 3 + q) x 2 (p 3 2q) x + (p 3 + q) = 0(C) (p 3 q) x 2 (5p 3 2q) x + (p 3 q) = 0 (D) (p 3 q) x 2 (5p 3 + 2q) x + (p 3 q) = 0

Key. (B)

Sol. 2 2 ++ =

3 + 3 = ( + ) {( + )2 3}q = p (p 2 3)

q + p 3 = 3 p =3(q p )

3p+

2 + 2 = p 2 23(q p )

3p+

=3 3 33p 2q 2p p 2q

3p 3p =

2 2 3

3

p 2qq p

+ = +

x2 3

3

(p 2q)x 1

p q ++

= 0 (p3 + q)x 2 (p 3 2q) x + (p 3 + q) = 0


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32. Equation of the plane containing the straight linex y z2 3 4

= = and perpendicular to the plane containing the

straight linesx y z3 4 2

= = and x y z4 2 3

= = is

(A) x + 2y 2z = 0 (B) 3x + 2y 2z = 0(C) x 2y + z = 0 (D) 5x + 2y 4z = 0

Key. (C)

Sol. n (3i 4 j 2k) (4i 2 j 3k)= + + + + = i j k

3 4 2

4 2 3

= 8i j 10k

The equation of plane containing the IInd and IIIrd given lines. r.(8i j 10k) 0 = 8x y 10z = 0.

Now normal vector to the required plane is given by

i j k

2 3 4

8 1 10 = 26i 52j 26k +

= 26(i 2 j k + )The equation of the required plane is x 2y + z = 0.

33. If the angle A, B and C of the triangle are in the an arithmetic progression and if a, b and c denote thelengths of the sides opposite to A, B and C respectively, then the value of the expressiona c

sin 2C sin 2Ac a

+ is

(A)12

(B)3

2

(C) 1 (D) 3 Key. (D)

Sol.

a csin 2C sin 2A

c a+

=

sin A sin C2sin C cos C 2sin A cos A

sin C sin A+

= 2sin(A + C)

= 2 3

2= 3

34. Let f, g and h be real valued functions defined on the interval [0, 1] by f(x) =2 2x xe e + , g(x) = xxe

2

and

h(x) = x 2 + xe2

. If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then(A) a = b and c b (B) a = c and a b(C) a b (D) a = b = c

Key. (D)Sol. 1 x x2 x[0, 1]

2x x 2 xe xe x e 2 2 x[0, 1]i.e.,

2x x x x x 2 xe e e xe e x e + + +2 2 2 2 2

equality holds when x = 1i.e., f(x) g(x) h(x) x[0, 1]Hence a = b = c.


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35. Let be a complex cube root of unity with 1. A fair die is thrown three times. If r 1 , r 2 and r 3 are thenumbers obtained on the die, then the probability that 31 2 rr r + + = 0 is

(A)1

18(B)

19

(C)29

(D)1

36

Key. (C)

Sol. Required prob. =2 2 2(3!) 2

6 6 6 9 =

29

36. Let P, Q, R and S be the points on the plane with position vectors 2 i j, 4i,3i 3j + and 3i 2 j + respectively. The quadrilateral PQRS must be a(A) parallelogram, which is neither a rhombus nor a rectangle(B) square (C) rectangle, but not a square(D) rhombus, but not a square

Key. (A)

Sol.

P : 2i j, Q:4i,R :3i 3j, S : 3i 2 j + + PQ = of P = 6i j+

QR 3i 3j 4i= + = i 3j + PS 3i 2 j 2i j= + + += i 3j +

SR 3i 3j 3i 2 j= + + = 6i j+

( ) ( ) PQ. PS 6i j . i 3j 3= + + =

0

Here PQ ||SR

and PS ||QR

but PQ is not perpendicular to PS

O

RS

2i j 4 i

3i 2 j + 3i 3j+

P Q



37. Let z 1 and z 2 be two distinct complex numbers and let z = (1 t) z 1 + tz 2 for some real number t with 0 < t 1 such that |f (x)| < |f(x)| for all x (, )(D) there exists > 0 such that |f(x)| + |f (x)| for all x (0, )

Key. (B, C)

Sol. f(x) = lnx +x

0

1 sin t dt+ , x > 0

f (x) = 1 1 sin xx

+ + , x > 0

Clearly f (x) exists for all x (0, )and f (x) is continuous on (0, )but not differentiable on (0, )More over f (x), f(x) > 0 x(1, )

and ln x +x

0

11 sin t dt 1 sin x

x+ > + + x(, )

1x

is not bounded.

(D) is incorrect .Hence, option B, C are correct.

40. The value(s) of 1 4 4

20

x (1 x)1 x

+ dx is (are)

(A)227

(B) 2105

(C) 0 (D)71 315 2

Key. (A)

Sol. 1 4 4

20

x (1 x)(1 x )

+ dx

1 1 46 5 4

20 0

4x(x 4x 5x )dx dx

1 x +

+ =1 4

20

10 x 1 14 dx

21 x 1 +

+ =227


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41. Let ABC be a triangle such that ACB =6

and let a, b and c denote the lengths of the sides opposite to A,

B and C respectively. The value(s) of x for which a = x 2 + x + 1, b = x 2 1 and c = 2x + 1 is (are)

(A) ( )2 3 + (B) 1 3+ (C) 2 3+ (D) 4 3

Key. (B)

Sol. 2 2 2 2 2

2 2

3 (x x 1) (x 1) (2x 1)2 2.(x 1) (x x 1)

+ + + += + +

32

= 2 2 2 2

2 2

(x 3x 2)(x x) (x 1)2(x 1)(x x 1)

+ + + + +

2

2

3 (x 2) x x 12 2(x x 1)

+ + =+ +

2

2

2x 2x 13

x x 1+ =+ +

A

B C

2x+1

x2 -1

x2 + x + 1

/6

x2 ( ) ( )3 2 x 3 2 3 1 0 + + + =

x =2( 3 2) ( 3 2) 4( 3 2) ( 3 1)

2( 3 2)

+ = 3 1+ .




Paragraph for Quest ions Nos . 42 to 44

Let p be an odd prime number and T p be the following set of 2 2 matrices.

Tp =a b

A :a, b, c {0, 1, 2, ..., p 1}c a

=

Sol. 42 to 44as A is symmetric b = c

det A = a 2 b 2 = (a + b) (a b)a, b, c, {0, 1, 2, .. p 1}no. of numbers of type

np = 1np + 1 = 1np + 2 = 1 n I

np + (p 1) = 142. The number of A in T p such that A is either symmetric or skew symmetric or both, and det(A) divisible by

p is(A) (p 1) 2 (B) 2 (p 1)(C) (p 1) 2 + 1 (D) 2p 1

Key. (D)Sol. as det(A) is div. by p either a + b div. by p corresponding nu. Of ways = (p 1) [excluding zero]

or (a b) is div. by p corresponding no. of ways = pTotal number of ways = 2p 1


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43. The number of A in T p such that the trace of A is not divisible by p but det (A) is divisible by p is[Note: The trace of a matrix is the sum of its diagonal entries.](A) (p q) (p 2 p p + q) (B) p 3 (p 1) 2 (C) (p 1) 2 (D) (p 1) (p 2 2)

Key . (C)Sol. as Tr(A) not div. by p a 0

det(A) is div. by p a2 bc div. by p

no. of ways of selection of a, b, c(p 1)[(p 1) 1] = (p 1) 2

44. The number of A in T p such that det (A) is not divisible by p is(A) 2p 2 (B) p 3 5p(C) p 3 3p (D) p 3 p 2

Key . (D)Sol. Total number of A = p p p = p 3

No. of A such that det(A) div. by p= (p 1) 2 + no. of A in which a = 0= (p 1) 2 + p + p 1= p 2 required no. = p 3 p 2.

Paragraph for Quest ions Nos . 45 to 46

The circle x 2 + y 2 8x = 0 and hyperbola2 2x y

19 4

= intersect at the points A and B.

45. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is(A) 2x 5 y 20 = 0 (B) 2x 5 y + 4 = 0(C) 3x 4y + 8 = 0 (D) 4x 3y + 4 = 0

Key . (B)Sol. Equation of tangent at point P( )

x sec y tan1 0

3 2 = ..(i)

since eq. (i) will be a tangent to the circle

2 2

4sec1

3 4sec tan

9 4

=

+

by solving it we will get

2x 5y 4 0 + =

(4,0)(3,0)(3,0)

(6, 2 3)

(6, 2 3)-

A

B

46. Equation of the circle with AB as its diameter is(A) x 2 + y 2 12x + 24 = 0 (B) x 2 + y 2 + 12x + 24 = 0(C) x 2 + y 2 + 24x 12 = 0 (D) x 2 + y 2 24x 12 = 0

Key. (A)

Sol. 9 2x ( x 8x)

19 4

+= +

4x2

= 36 + 9( x2

+ 8x)13x 2 72x 36 = 0x = 6,y 2 3= Required equation of circle is(x 6) 2 + y 2 12 = 0x2 + y 2 12x + 24 = 0

(4,0)(3,0)(3,0)

(6, 2 3)

(6, 2 3)-

A

B


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47. Let be the complex number cos 2 2isin3 3

+ . Then the number of dist inct complex numbers z satisfying

2

2

2

z 1z 1

1 z

+ + +

= 0 is equal to

Key. (1)

Sol. 2 2 1 3

cos i sin i3 3 2 2 = + = +

is one of cube root of unity.2

2

2

z 1

z 1 0

1 z

+ + = +

R1 R1 + R 2 + R 3

2 2

2

z z z

z 1 0 1 0

1 z

+ = + + = +

C1 C1 C 2 & C 2 C2 C 3 gives

2 2

2

0 0 z

z z 1 1 0

1 1 z z

+ = +

2 2 2z ( z )(1 z ) ( 1)(z 1) 0 + = z[z 2] = 0

3

z 0 = = z = 0

Ans. is = 1

48. The number of values of in the interval ,2 2

such that n

5

for n = 0, 1, 2 and tan = cot5 as

well as sin2 = cos4 isKey. (3)Sol. tan cot 5 =

tan tan 52 =

n 0 52

= +

6 n2 = +

nn I

6 12 = + ..(i)

sin 2 cos 4 =


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51. The line 2x + y = 1 is tangent to the hyperbola2 2

2 2

x y1

a b = . If this line passes through the point of

intersection of the nearest directrix and the x axis, then the eccentricity of the hyperbola isKey. (2)

Sol. Since the line 2x + y 1 = 0 is tangent

so, C 2 = a 2m2 b 2 1 = 4a 2 b 2 ..(i)

Also line passes througha

,0e

So,a

2 1e

=

4a 2 = e 2 (ii)Using (i) and (ii) e = 2

(0, 0)

ax

e=

52. Let S k , k = 1, 2, ... , 100, denote the sum of the infinite geometric series whose whose first term isk 1

k!

and the common ratio is1k

. Then the value of ( )2 100

2k

k 1

100k 3k 1 S

100! =+ + is

Key. (4)

Sol. k

k 11k!S

1 (k 1)!1k

= =

We have S 1 = 1S2 = 1

S3 =12

Now, ( )100

2k

k 1

k 3k 1 S=

+

= S 1 + S 2 + S 3 +( )

( )

2100

k 4

k 3k 1

k 1 !=

+

= 1 + 1 +( ) ( )

100

k 4

1 1 12 k 3 ! k 1 !=

+

= 1 + 1 +1 1 1 1

12 2! 98! 99! + +

= 4 -10099!

So, ( )2 100

2k

k 1

100k 3k 1 S 4

100! =+ + = .

53. Let f be a real valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P,then the value of f( 3) is equal to

Key. (9)


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Sol. eq. of tangent at P(x, y)

dyY y (X x )

dx =

yinteger 3dy

y x xdx

=

2dy yxdx x =

I.F. =1

dxx 1e

x

=

The solution

21 1y x dxx x

= 2y x

Cx 2

= +

3f (1) 1 C

2= =

33x xf (x) y 2

= =

f ( 3) 9 =

54. If a and b are vectors in space given by i 2 j

a5

= and 2i j 3k

b14

+ += , then the value of

( ) ( ) ( )2a b . a b a 2b +

is

Key. (5)

Sol. | a | | b | 1 a.b 0= = = Let (a b) (a 2b) (a b) a 2(a b) b= =

l

2| a | b (a.b)a 2(a.b)b 2 | b | a.= +

b 2a= + 2(2a b). | 2a b | 5+ = + =

l

55. The number of all possible values of , where 0 < < , for which the system of equations(y + z) cos3 = (xyz) sin 3

xsin3 = 2 cos3 2sin 3y z

+

(xyz) sin3 = (y + 2z) cos3 + ysin 3 have a solution (x 0, y0, z0) with y 0 z0 0, isKey. (3)

Sol. (y z) cos 3 (xyz)sin 3+ = (A)2cos3 2sin 3

xsin3y z

= + (B)

(xyz)sin3 (y 2z)cos3 ysin 3 = + + (C)When cos 3 0.


19/57



y z 2z y 2ztan3

xyz y(xz 2) xyz y+ + = = =

as y 0(y + z) (xz 2) = 2z(xz)xyz + xz 2 2z 2y = 2xz 2 xyz = 2y + 2z + xz 2 ..(i)

Again, 2z(xz 1) = (y + 2z) (xz 2)2xz 2 2z = xyz + (2xz 2 4z 2y)xyz = 2y + 2z ..(ii)

from (i) and (ii) xz 2 = 0 x = 0 as z 0from (A) (y + z) cos 3 = 0 y + z = 0But when cos 3 = 0 from (B)

sin 3 = 0 not possibleSo y = z putting in (B) and (C)

x = 0sin 3 = cos 3

tan3 = 1 5 3, ,12 12 4 =

56. The maximum value of the expression 2 21

sin 3sin cos 5cos + + is

Key. (2)

Sol. Let 2 21 1

y3sin 3sin cos 5cos 3 2cos 2 sin 22

= = + + + +

5 3 52cos 2 sin 2

2 2 2 +

max. value of 1

y 25

32

= =


20/57



PART III : PHYSICS



57. An AC voltage source of variable angular frequency and fixed amplitude V 0 is connected in series with acapacitance C and an electric bulb of resistance R (inductance zero). When is increased(A) the bulb glows dimmer (B) the bulb glows brighter(C) total impedance of the circuit is unchanged (D) total impedance of the circuit increases.

Key. (B)

Sol. P = V rms Irms cos 2

rmsV RZ Z

= 2

rms2

V RZ

=

2

2 2

1Z R C= +

As increase Z, decreases, so P increases.Hence correct option is (B).

58. A thin flexible wire of length L is connected to two adjacent fixedpoints and carries a current I in the clockwise direction, as shown inthe figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shapeof a circle. The tension in the wire is

(A) IBL (B)IBL

(C)IBL2

(D)IBL4

.

Key. (C)

Sol. 2T iB(2R)= T

iB(2R)

T

iBL

T2

=

Hence correct option is (C).

59. A block of mass m is on an inclined plane of angle . The coefficient of friction between the block and the plane is and tan > . The block is heldstationary by applying a force P parallel to the plane. The direction of forcepointing up the plane is taken to be positive. As P is varied from P 1 = mg (sin cos ) to P 2 mg (sin + cos ), the frictional force f versus P graph

P


21/57



will look like

(A) (B)f

PP1

P2

f

PP1 P2

(C) (D)

f

PP1

P2

f

PP1 P2

.Key. (A)Sol. 1P mg(sin cos )=

friction initial = mgcos up along the planefriction final = mgcos down along the plane

Hence correct option is (A).

60. A real gas behaves like an ideal gas if its(A) pressure and temperature are both high (B) pressure and temperature are both low(C) pressure is high and temperature is low (D) pressure is low and temperature is high.

Key. (D)

Sol. For ideal gas behaviour pressure should be low and temperature should be high.Hence correct option is (D).

61. Consider a thin square sheet of side L and thickness t, made of a material of resistivity . The resistance between two opposite faces, shown by the shadedareas in the figure is(A) directly proportional to L(B) directly proportional to t(C) independent of L(D) independent of t.

t

L

Key. (C)

Sol. L

RA Lt t

= = =


62. A thin uniform annular disc (see figure) of mass M has outer radius 4R andinner radius 3R. The work required to take a unit mass from point P on itsaxis to infinity is

(A) ( )2GM 4 2 57R (B) ( )2GM

4 2 57R

(C)

GM

4R (D) ( )2GM

2 15R .

3R 4R

P4R

Key. (A)


22/57



Sol.

4R

r x

3R

2 2

G 2 rdrdV

r x

= +

4R

2 23R

rdrV 2 G

r x=

+ 2 2r x z+ =

2r dr dz=

2 2

rdr dz

2 zr x=

+ 1 z

z122= =

4R2

3RV 2 G r x2 = +

2 Gr 4R 2 5 =

W (1) 0 2 G (4R 2 5R) = +

2

M2 G (4R 2 5R)

(16 9)R=

2 GM(4 2 5)

7R= .


63. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with theincrease in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistancesR100 , R 60 and R 40, respectively, the relation between these resistances is

(A)100 40 60

1 1 1R R R

= + (B) 100 40 60R R R= +

(C) 100 60 40R R R> > (D)100 60 40

1 1 1R R R

> > .

Key. (D)

Sol.2v

RP

=

As temperature increase, resistance increasesSo, 40 60 100R R R> > .Hence correct option is (D).


23/57



64. To verify Ohm's law, a student is provided with a test resistor R T, a high resistance R 1, a small resistanceR2, two identical galvanometers G 1 and G 2, and a variable voltage source V. The correct circuit to carry outthe experiment is(A) (B)

G1

V

G2R 2

R1RT

G1

V

G2R 2

R 1

RT

(C) (D)G1

V

G 2

R 2

R 1

RT

G1

V

G 2R 2

R 1RT

.Key. (C)

Sol. An ideal voltmeter should have large resistance and an ideal ammeter should have low resistance.Hence correct option is (C).



65. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kgmass reverses its direction and moves with a speed of 2 ms 1. Which of the following statement(s) is (are)correct for the system of these two masses ?(A) total momentum of the system is 3 kg ms 1

(B) momentum of 5 kg mass after collision is 4 kg ms 1 (C) kinetic energy of the center of mass is 0.75 J

(D) total kinetic energy of the system is 4 J.Key. (A), (C)Sol. (1) (V) + (5) (0) = (1) (2) + 5 V

V = 5V 2 (i)V ' 2

1V 0

+ =

V = V 2 (ii)V = 5 (V 2) 2

From equation (i) and (ii)V = 5V 10 24V = 12V = 3 m/s.Pi = (1) (3) = 3 kg m/s.

CM

(1)(3) (5)(0) 1V m / s6 2

+= =

CM

1 1 3K (6) 0.75 J

2 4 4= = =

2total

1K (1)(3) 4.5 J

2= =

Hence correct options are (A), (C).


24/57



66. A few electric field lines for a system of two charges Q 1 and Q 2 fixedat two different points on the xaxis are shown in the figure. Theselines suggest that(A) |Q 1| > |Q 2|(B) |Q 1| < |Q 2|(C) at a finite distance to the left of Q 1 the electric field is zero(D) at a finite distance to the right of Q 2 the electric field is zero.

Q1 Q2

Key. (A), (D)Sol. Density of field lines is more are Q 1

|Q1| > |Q 2|Q1 and Q 2 are of opposite signsSo, null point will be closer to charge of smaller magnitude i.e., Q 2 Hence correct options are (A), (D).

67. A ray OP of monochromatic light is incident on the face AB of prism ABCDnear vertex B at an incident angle of 60 (see figure). If the refractive index of the material of the prism is 3 , which of the following is (are) correct ?(A) the ray gets totally internally reflected at face CD(B) the ray comes out through face AD

(C) the angle between the incident ray and the emergent ray is 90(D) the angle between the incident ray and the emergent ray is 120.

O B

C

A D

P60

135

9075

Key. (A), (B), (C)Sol. 1sin 60 3 sin r=

r = 30

C

1sin

3 =

C 35 B

C

A D

75

6030

60

6030

4545

45

At CD angle of incidence is greater than C .At AD angle of incidence is less than critical angleSo ray will come out of AD.Angle of deviation

30 + 90 + 30 = 90Hence correct options are (A), (B), (C)

68. One mole of an ideal gas in initial state A undergoes a cyclic processABCA, as shown in figure. Its pressure at A is P 0. Choose the correct

option(s) from the following :(A) internal energies at A and B are the same(B) work done by the gas is process AB is

0 0P V n 4

(C) pressure at C is 0P4

VB

AC

T

4V 0

V0

T0

(D) temperature at C is 0T4

.


25/57



Key. (A), (B), (C), (D)

Sol. Internal energy of an ideal gas depends on temperature

2BC

1

VW nRT n

V=

0 0 0

0

P V 4V(1)(R) n

R V

=

0 0P V n4= For CA

Pconstant

T=

P at C 0P4

=

T at C 0T

4=

Hence all options are correct.

69. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. Heuses a stop watch with the least count of 1 second fore this and records 40 seconds for 20 oscillations. Forthis observation, which of the following statement(s) is (are) true ?(A) error T in measuring T, the time period, is 0.05 seconds(B) error T in measuring T, the time period, is 1 second(C) percentage error in the determination of g is 5%(D) percentage error in the determination of g is 2.5%.

Key. (A), (C)

Sol. Error in measurement of 1

T s 0.05 s20

= =

dg dT2

g T=

dg 12

g 40=

% error in calculation of g = 5%.




Paragraph for Quest ion Nos . 70 to 72When a particle of mass m moves on the xaxis in a potential of the form V (x) = kx 2, it performs simple

harmonic motion. The corresponding time period is proportional tomk

, as can be seen easily using

dimensional analysis. However, the motion of a particle can be periodic even when its potential energy

increases on both sides of x = 0 in a way different from kx 2 and its total energy is such that the particle doesnot escape to infinity. Consider a particle of mass m moving on the xaxis. Its potential energy is V(x) =x4 ( > 0) for |x| near the origin and becomes a constant equal to V 0 for |x| X0 (see figure).

V0

X0x

V(x)


26/57



70. If the total energy of the particle is E, it will perform periodic motion only if

(A) E < 0 (B) E > 0(C) V 0 > E > 0 (D) E > V 0.

Key. (B)Sol. For periodic motion

Total energy should be less than V 0 but greater than zero.

Hence (C) is correct.

71. For periodic motion of small amplitude A, the time period T of this particle is proportional to

(A)m

A

(B)1 mA

(C) Am

(D)1A m

.

Key. (B)Sol. Dimensionally only B is correct.

72. The acceleration of this particle for |x| > X 0 is

(A) proportional to V 0 (B) proportional to 00

VmX

(C) proportional to 00

VmX

(D) zero.

Key. (D)Sol. For x > x 0

potential energy is constantforce on particle is zero.Hence (D) is correct.

Paragraph for Quest ion Nos . 73 to 74Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero valueto zero as their temperature is lowered below a critical temperature T C(0). An interesting property of superconductors is that their critical temperature becomes smaller than T C(0) if they are placed in magnetic

field, i.e., the critical temperature T C(B) is a function of the magnetic field strength B. The dependence of TC(B) on B is shown in the figure.

T (B)C

T (O)C

O B 73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for

two different magnetic field B 1 (solid line) and B 2 (dashed line). If B 2 is larger than B 1, which of thefollowing graphs shows the correct variation of R with T in these fields ?(A)

O

R

TB2 B1

(B)

O

R

T

B2

B1

(C)

O

R

TB2B1

(D)

O

R

TB2

B1


27/57



Key. (A)Sol. As B increases, critical temperature decreases.

74. A superconductors has T C(0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its T C decreases to 75K. For this material one can definitely say that when(A) B = 5 Tesla, T C(B) = 80 K (B) B = 5 Tesla, 75 K < T C(B) < 100 K

(C) B = 10 Tesla, 75 K < T C(B) < 100 K (D) B = 10 Tesla, T C(B) = 70 K.Key. (B)



75. A binary star consists of two starts A (mass 2.2. M S) and B (mass 1 M S), where M S is the mass of the sun.They are separated by distance d and are rotating about their center of mass, which is stationary. The ratioof the total angular momentum of the binary star to the angular momentum of star B about the center of mass is.

Key. 6.Sol.

2.2M S

A B

r1 r2

11M S 2 22 1

22

11r 2.2 r11r

+

2 111 r 2.2 r= 2

122

r2.21

11 r= +

22.2 11

1 611 2.2

= + =

.

76. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in

front of it to 50 cm, the magnification of its image changes from m 25 to m 50. The ratio 2550

mm

is

Key. 6.

Sol. | f |

m| f u |

=

25

50

m6

m= .

77. A 0.1 kg mass is suspended form a wire of negligible mass. The length of the wire is 1 m and its crosssectional area is 4.9 10 7 m2. If the mass is pulled a little in the vertically downward direction andreleased, it performs simple harmonic motion of angular frequency 140 rad s 1. If the Young's modulus of

the material of the wire is n 109

Nm2

, the value of n isKey. 4.

Sol. 2Km

= 7YA Y(4.9 10 )

140 140m (1)(0.1)

= =

7140 140 y(49) 10 =


28/57



9y 4 10= n = 4.

78. When two progressive waves y 1 4 sin (2x 6t) and y 2 = 3 sin 2x 6t2

are superimposed, the

amplitude of the resultant wave is

Key. 5.Sol. Amplitude 2 24 3 5= + = .

79. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T 1 and T 2, respectively. Themaximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Consideringthem to be black bodies, what will be the ratio of total energy radiated by A to that of B ?

Key. 9.Sol. (T1) (500 nm) = T 2 (1500 nm)

T1 = 3T 2 2 4

A 1E 4 (6cm) (T )= 2 4

B 1E 4 (18cm) (T )= 2

4A

B

E 1(3) 9

E 3 = =

.

80. Gravitational acceleration on the surface of a planet is6

g11

, where g is the gravitational acceleration on

the surface of the earth. The average mass density of the planet is23

times that of the earth. If the escape

speed on the surface of the earth is taken to be 11 kms 1, the escape speed on the surface of the planet inkms 1 will be

Key. 2.

Sol. eP2 2P e

GMGM 6 6g

11 11R R= = (i)

e e2g R 11 km / s=

P P2g R x= 2

e e2

P P

g R (11)g R x

=

ee2

e2

PP2

P

GMR

R 21GM xRR

1=

e

e2

P

P

MR 121M xR

= (ii)

eP3 3P e

MM 23R R

= (iii)

x = 2.

81. A stationary source is emitting sound at a fixed frequency f 0, which is reflected by two cars approaching thesource. The difference between the frequencies of sound reflected from the cars is 1.2% of f 0. What is thedifference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constantspeeds much smaller than the speed of sound which is 330 ms 1.


29/57



Key. 7.

Sol. 1

1

C1 0

C

V Vf f

V V

+=

; 2

2

C2 0

C

V Vf f

V V

+=

1 2

1 2

C C0 0

C C

V V V V 1.2f f f

V V V V 100

+ + = =

C 002 V 1.2f f V 100= =

CV 7 km / hr = .

82. When two identical batteries of internal resistance 1 each are connected in series across a resistor R, therate of heat produced in R is J 1. When the same batteries are connected in parallel across R, the rate is J 2. If J1 = 2.25 J 2 then the value of R in is

Key. 4.

Sol. 2

1

2J R

2 R = +

2

2J R0.5 R = +

2

24(0.5 R)2.25

(2 R)+= +

9 4(R 0.5)4 2 R

+=+

3 2R 22 2 R

+=+

6 3R 4R 2+ = + R 4= .

83. A piece of ice (heat capacity = 2100 J kg 1 C 1 and latent heat = 3.36 10 5 J kg 1) of mass m grams is at 5C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the icewater mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m is

Key. 8.Sol. 5 3m(2100)(5) 1(3.36 10 ) 10 420 + =

11m 336 420+ = 11m 420 336= 84= m 8 gm= .

84. An particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de

Broglie wavelengths are pand respectively. The ratiop

, to the nearest integer, is

Key. 3.

Sol. h

2mk =

k qV= h

2(m)qV =

p

p p

m q (4m)(2q)2 2 3

m q (m)q

= = = =

.


30/57



IIT JEE (2010) PAPER II QUESTION & SOLUTIONS CODE 0 PART I : CHEMISTRY

PAPER II



1. The compounds P, Q and S

HO

COOH

CH3

OCH 3 CO

O

P Q S where separately subjected to nitration using HNO 3 /H 2SO 4 mixture. The major product formed in each caserespectively, is

(A)HO

COOH

NO 2

CH3

OCH 3

NO 2

CO

O

O2N

(B)HO

COOH

NO 2 CH3

OCH 3

NO 2

CO

O

NO 2

(C)

HO

COOH

NO 2

CH3

OCH 3

NO 2

CO

ONO 2

(D)

HO

COOH

NO 2

CH3

OCH 3

NO 2

CO

O

NO 2

Key: (C)

Sol.:

OH

COOH

OH

NO2

COOH

P


31/57



CH3

OCH 3

CH3

OCH 3

NO 2

Q

S C

O

O C

O

O NO 2

2. Assuming that Hunds rule is violated, the bond order and magnetic nature of the diatomic molecule B 2 is

(A) 1 and diamagnetic (B) 0 and diamagnetic(C) 1 and paramagnetic (D) 0 and paramagnetic.

Key: (A)

Sol.: 2B 10 electron

x y

2 2 2 2 1 11S 1S 2S 2S 2p 2p* * (If Hunds rule is obeyed)

x y

2 2 2 2 2 01S 1S 2S 2S 2p 2p* * (If Hunds rule is violated)

Bond order =6 4

1

2

=

Paramagnetic if Hunds rule is obeyedDiamagnetic of Hunds rule is violated.

3. The packing efficiency of the two-dimensional square unit cell shown belowis(A) 39.27% (B) 68.02%(C) 74.05% (D) 78.54%.

L Key: (D)Sol.: Let us consider the squerare plane (given)

Area = L2

Area covered by 22 r (two circle)

4r 2= L2

r4

= L

Area =2

222 4 L2 L L

4 16 4

= =

Packing efficiency2

2

L100

4 L

=

3.14100

4= = 78.5%.

4. The complex showing a spin-only magnetic moment of 2.82 B.M. is

(A) ( )4Ni CO (B) [ ]2

4NiCl

(C) ( )3 4Ni PPh (D) ( )2

4Ni CN

Key: (B)


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Sol.: 2 828Ni 4S 3d2 0 8Ni 4S 3d+

3d 4s 4p

2 unpaired electron (n = 2)( )n n 2 = + 8= BM

= 2.82 BM.

5. In the reaction CH3 C

NH 2

O

C

Cl

O

NaOH/Br 2T

(1)

(2)

the structure of the Product T is

(A)

C

O

O

CH3C

O

(B)NH

C

O

CH 3

(C)NH

C

O

CH3

(D)

C

NH

O

CH3C

O

Key: (C)

Sol.:

C

O

NH 2

CH3

NH2

CH3

NH

CH3

C

OC

O

Cl

( ) 21 NaOH/BrHoffman Bromide reaction Acylation

(2)

6. The species having pyramidal shape is(A) 3SO (B) 3BrF

(C) 23SiO (D) 2OSF

Key: (D)

Sol.:S

O FF

Pyramidal


33/57



SECTION IIInteger Type

This section contains a group of 5 questions. The answer to each of the questions is a single-digit integer.ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled.

7. Silver (atomic weight = 108 g mol 1) has a density of 10.5 g cm 3. The number of silver atoms on a surfaceof area 10 12 m2 can be expressed in scientific notation as xy 10 . The value of x is

Key: (7)Sol.: Consider a single layer square shaped arrangement of n n silver atoms. Also assume the radius of each

silver atom r.Area of the layer = (2rn) 2= 10 -12m2 = 10 -8cm2 Mass of the layerSurface area thickness density = number of atoms mass of single atom.

8 2 2410 2r 10.5 n 108 1.66 10 = [put 2rn = 10 4] 3 10n 5.855 10=

3n 3.82 10 number of silver atoms on the surface

( )2

2 3n 3.82 10= = = 1.4592 107

x = 7.

8. Among the following, the number of elements showing only one non-zero oxidation state is

O, Cl, F, N, P, Sn, Tl, Na, TiKey: (2)Sol.: Fluorine and sodium shown only one non zero oxidation state, fluorine show 1 and sodium shown + 1.

9. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines asshown in the graph below. If the work done along the solid line path is w s and that along the dotted linepath is w d, then the integer closest to the ratio w d /w s is

0.0 0.5 1.0 1.5 2.0 2.5 3.0 4.0

4.5

5.0 5.5 6.00.0

0.51.0

1.52.02.53.0

3.5

3.54.0

4.5

P

b

(atm.)

V(lit.)

a

Key: (2)

Sol.: wd = 4 1.5 + 1 15 + 0.80 2.5= 9.375

ws = 2.33 pV log 21

vv

= 2.33 4 0.5 log 5.50.5

4.606


34/57



ds

w 9.3752

w 4.606= .

10. The total number of diprotic acids among the following is

H3PO 4 H2SO 4 H3PO 3 H2CO 3 H2S2O7

H3BO 3 H3PO 2 H2CrO 4 H2SO 3 Key: (6)Sol.: The total number of diprotic acids are 6

O

S

OOH OH

O

P

OHOH H

O

COH OH

O

S

OOH

O S OH

O

O

O

Cr

OOH OH

O

SOH OH

11. Total number of geometrical isomers for the complex [RhCl(CO)(PPh 3)(NH 3)] isKey: (3)Sol.: The total number of geometrical isomers for the complex [RhCl(O)(PPh 3)(NH 3)] is 3.

M

C

A

D

B

M

D

A

C

B

M

B

A

D

C

SECTI ON II IParagraph Type

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 12 to 14 Two aliphatic aldehydes P and Q react in the presence of aqueous K 2CO 3 to give compound R, which upontreatment with HCN provides compound S. On acidification and heating, S gives the product shown below :

O O

OHCH3

CH3

12. The compounds P and O respectively are :

(A)CH3

CH

C

H

O

CH3

and

CH3C

H

O

(B)CH3

CH

C

H

O

CH3

and

H

C

H

O

(C) and

CH3C

H

O

C

H

O

CH 2CH3

CH

CH3

(D) and

H

C

H

O

C

H

O

CH 2CH3

CH

CH3 Key: (B)


35/57



13. The compound R is

(A)

CH3

CC

H

O

CH 2OH

H3C

(B) CH3C

CH

OHCH3

CH3C

O

H

(C)CH3

CH CH

C

CH 2

CH3

OH

O

H

(D)CH3

CH

CH

CH

CH 3

CH3 OH

C

O

H

Key: (A)

14. The compounds S is

(A)CH

CH 2

CN

CH3

CH

CCH 3

O

H

(B) CH3C

H2CCN

CH3 C

O

H

(C)CH3

CH CH

CH

CH2

CH3

OH

CN

OH

(D) CH3C

CH 2

OH

CH3CH

CN

OH

14. (D)

Sol.:(12-14)

CH3

C

C

H

O

CH3H

H

C

H

O

(P) (Q)

K2CO 3

CH3

C

C H

O

CH3

H

C OH


36/57



O

CH3 C C

CH3

CH3

CH 2 OH

(R)

OH

CH3 C C

CH3

CH3

CH 2 OH

CN

(S)

OH

CH3 C C

CH3

CH3

CH 2 OH

CO 2H

O

OH

O

Cyclic ester (lactone)

HCN 3H O

Paragraph for Questions 15 to 17

The hydrogen-like species Li 2+ is in a spherically symmetric state S 1 with one radial node. Upon absorbinglight the ion undergoes transition to a state S 2. The state S 2 has one radial node and its energy is equal to theground state energy of the hydrogen atom.

15. The state S 1 is(A) 1s (B) 2s(C) 2p (D) 3s

Key: (B)Sol.: No. of radial node = n 1

Since state S 1 has 1 radial node it must be 2s orbital with n = 2 and 0= . (B)

16. Energy of the state S 1 in units of the hydrogen atom ground state energy is(A) 0.75 (B) 1.50(C) 2.25 (D) 4.50

Key: (C)

Sol.: Energy of state S 1 = 2213.6

32

eV/atom

= 13.6 2.25 eV / atom= 2.25 times energy of ground state of hydrogen atom.

(C)

17. The orbital angular momentum quantum number of the state S 2 is(A) 0 (B) 1(C) 2 (D) 3

Key: (B)Sol.: Energy of S 2 level = 13.6 eV/atom

2

2

13.6 313.6

n =

P.Q.N of level S 2 = 3Since S 2 has one radial node being present in 3

rd shell it must be 3p orbital.

The orbital angular momentum quantum numberi.e., Azimuthal quantum number of S 2 = 1.


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SECTION IVMatrix Type

This section contains 2 questions . Each question four statements (A, B, C and D) given in Column I and fivestatements (p, q, r, s and t) in Column II . Any given statement in Column I can have correct matching with one or more statement(s) given in Column II . For example, if for a given question, statement B matches with thestatements given in q and r, then for that particular question, against statement B, darken the bubbles

corresponding to q and r in the ORS.

18. Match the reactions in Column I with appropriate options in Column II .Column I Column II

(A)N2Cl OH

2NaOH/H O0 C N N OH

(p) Racemic mixture

(B)

CH3 C

CH3

OH

C CH 3

CH3

OH

2 4H SO CH3C

C CH3

CH3

CH3

O

(q) Addition reaction

(C)C

O

CH3

4

3

1. LiAlH

2. H O + CH

OH

CH3

(r) Substitution reaction

(D)ClHS

Base S

(s) Coupling reaction

(t) Carbonation intermediate

Key: (A r, s), (B t), (C p, r), (D r)

Sol.: (A) N2 OHH N N OH

(B) CH3 C

OH

CH3

C CH 3

OH

CH3H+ CH3 C

CH3

C CH 3

OH

CH3

CH3 C

CH3

CH3

C CH 3

O

(C) C

O

CH3 4LiAlH C

HO

CH3

H

C

O

CH3

H

AlH 3

2H O

(D) Intramolecular substitution reaction


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19. All the compounds listed in Column I react with water. Match the result of the respective reactions with theappropriate options listed in Column II .

Column I Column II

(A) ( )3 22CH SiCl (p) Hydrogen halide formation

(B) XeF 4 (q) Redox reaction

(C) Cl 2 (r) Reacts with glass

(D) VCl 5 (s) Polymerization

(t) O 2 formation

Key: (A p, s), (B p, q), (C p, q), (D p)

Sol.: (A) CH3 Si

H

CH3

Cl 22H O CH3 Si

OH

CH3

OH Polymerisation

( )silicons

(B)

2XeF 4 + 3H 2O Xe + XeO 3 + 3H 2F2 + F 2

SiO 2 + 4HF SiF 4 + 2H 2O

SiF 4 + 2HF H2SiF 6

(C) Cl2 + H 2O HOCl + HCl.

(D) 5 2VCl H O+ must form HCl on hydrolysis


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PART II: MA THEMAT ICS


This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of

which ONLY ONE is correct.

20. If the distance of the point P(1, 2, 1) from the plane x + 2y 2z = , where > 0, is 5, then the foot of theperpendicular from P to the plane is

(A)8 4 7

, ,3 3 3

(B)4 4 1

, ,3 3 3

(C)1 2 10

, ,3 3 3

(D)2 1 5

, ,3 3 2

Key (A)

Sol. |5 + | = 15 = 10

If (x 1, y1, z1 ) is foot of perpendicular 1 1 1x 1 y 2 z 1 ( 15)

1 2 2 9

+ = = =

(x1, y1, z1) (8/3, 4/3, -7/3)

21. A signal which can be green or red with probability4 1

and5 5

respectively, is received by station A and then

transmitted to station B. The probability of each station receiving the signal correctly is34

. If the signal

received at station B is green, then the probability that the original signal was green is

(A)35

(B)67

(C)2023 (D)

920

Key (C) Sol. E1 original signal is green.

E2 original signal is red.E signal received at station B is green.

P(E 1 /E) = 1 11 1 2 2

p(E ) p(E) / E )p(E ) p(E / E ) p(E ) p(E / E )+

=

2 2

2 2

4 3 15 4 4

4 3 1 1 3 1 1 3

5 4 4 5 4 4 4 4

+

+ + +

=2023

22. Two adjacent sides of a parallelogram ABCD are given by AB 2i 10j 11k = + + and AD i 2j 2k = + + . Theside AD is rotated by an acute angle in the plane of the parallelogram so that AD becomes AD . If AD makes a right angle with the side AB, then the cosine of the angle is given by


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Key (B )Sol. f (f -1(x)) = x

f (f -1 (x)) . (f -1(x)) = 1

(f -1(x)) = 11

f (f (x))

We have to find (f -1(2))

(f -1(2)) = 11

f (f (2))

When f -1 (x) = 2 f(x) = 2 x = 0

given e -x f(x) = 2 +x

4

0

t 1+ dte-x (-f(x) + f (x)) = x 4 + 1put x = 0 f (0) = 3 (f -1 (2)) = 1/3.

25. Let S = {1, 2, 3, 4). The total number of unordered pairs of disjoint subsets of S is equal to(A) 25 (B) 34(C) 42 (D) 41

Key (D)Sol. S = {1, 2, 3,4}, then No. of unordered pairs of disjoint subsets of S is

43 141

2+ =



26. Let a 1, a2, a3, ... , a 11 be real numbers satisfying a 1 = 15, 27 2a 2 > 0 and a k = 2a k-1 a k-2 for k = 3, 4, ... , 11.

If 2 2 21 2 11a a ... a

11

+ + += 90, then the value of 1 2 11

a a ... a11

+ + +is equal to

Key (0)

Sol. a12 + a 2

2 + a 32 + .... + a 11

2 = 990 a2 + (a + d) 2 + (a + 2d) 2 + ... + (a + 10d) 2 = 990 11a 2 + d 2 (12 + 2 2 + 3 2 + ... + 10 2) + ad (2 + 4 + 6 + ... + 20) = 990 11 225 + d 2 385 + d 15 110 = 990 7d2 + 30d + 27 = 0 d = -3, -9/7 (n.p.) d = -3 and a 1 = 15

1 2 11a a ... a 11

(2 15 10 ( 3))

11 2 11

+ + + = +

= 0

27. Let f be a function defined on R (the set of all real numbers) such that f (x) = 2010 (x 2009) (x 2010) 2 (x 2011) 3 (x 2012) 4, for all x R.If g is a function defined on R with values in the interval (0, ) such that f(x) = ln (g(x)), for all x R, thenthe number of points in R at which g has a local maximum is

Key (1)


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Sol. The sign scheme of f (x)+++

2009 2010 2011 2012The local maximum of f(x) occurs at x = 2009Hence, local maximum of g(x) also occurs at x = 2009. Hence the number of point of local maximum = 1.

28. Let k be a positive real number and let A =

2k 1 2 k 2 k

2 k 1 2k

2 k 2k 1

and B =

0 2k 1 k

1 2k 0 2 k

k 2 k 0

. If

det (adj A) + det (adj B) = 10 6, then [k] is equal to{Note: adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equalto k].

Key (4) Sol. |A| = (2k 1) (-1 + 4k 2) + 2 k (2 k 4k k + ) + 2 k (4k k 2 k + )

(2k 1) (4k 2 1) + 4k + 8k 2 + 8k 2 + 4k = (2k 1) (4k 2 1) + 8k + 16k 2 = 8k 3 4k 2 2k + 1 + 8k + 16k 2 = 8k 3 + 12k 2 + 6k +1

|B| = 0 as B is skew symmetric matrix of odd order. (8k 3 + 12k 2 + 6k + 1) 2 = (10 3)2 (2k + 1) 3 = 10 3 2k + 1 = 10 k = 4.5[k] = 4.

29. Two parallel chords of a circle of radius 2 are at a distance 3 1+ apart. If the chords subtend at the center,

angles of 2

andk k

, where k > 0, then the value of [k] is

[Note : [k] denotes the largest integer less than or equal to k]Key (3)

Sol. 2cos 2cos 3 1k 2k + = +

2cos 2 3 3

cos2k 2k 2 ++ =

cos21 (2 3 1)

2k 4

+ = = 32

or3 12

But cos3 1

2k 2

cos3

cos2k 2 6 = = k = 3.

C r=2

/2k

/k

30. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and Crespectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3 . If ACB is obtuse and if r denotes

the radius of the incircle of the triangle, then r2

is equal toKey (3) Sol. The area = 15 3

1

6 102

sinC = 15 3

C = 120

c

B

AC b = 10

a= 6 r rr


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cosC =2 2 210 6 c2 10 6

+

-60 = 136 c 2 c2 = 196 c = 14.

Since r =15 3

s (10 6 14) / 2 =

+ +=

15 3 230

= 3

r2 = 3.



Paragraph for Quest ions Nos . 31 to 33Consider the polynomial f(x) = 1 + 2x + 3x 2 + 4x 3 Let s be the sum of all distinct real roots of f(x) and let t = |s|.

31. The real number s lies in the interval

(A)1

, 04

(B)3

11,4

(C) 3 1,4 2 (D) 10, 4

Key (C) Sol. f(x) is constant function

and3 1

f f 04 2

>


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35. The orthocenter of the triangle PAB is

(A)8

5,7

(B)7 25

,5 8

(C)11 8

,5 5

(D)8 7

,25 5

Key (C)

Sol. Equation of PEy 4 = 3(x 3) ..(i)Equation of AD

8y

5= ..(ii)

Solving (i) & (ii) we get11 8

x , y5 5

= =

36. The equation of the locus of the point whose distances from the point P and the line AB are equal, is(A) 9x 2 + y 2 6xy 54x 62y + 241 = 0 (B) x 2 + 9y 2 + 6xy 54x + 62y 241 = 0(C) 9x 2 + 9y 2 6xy 54x 62y 241 = 0 (D) x 2 + y 2 2xy + 27x + 31y 120 = 0

Key (A)

Sol.

22 2 ( 3 3)

( 3) ( 4) 10

+ + =

2 2 2 210( 6 9 8 16) 9 9 6 6 18 + + + = + + + Required locus is 9x 2 + y 2 6xy 54x 62y + 241 = 0


This section contains 2 questions . Each question four statements (A, B, C and D) given in Column I and fivestatements (p, q, r, s and t) in Column II . Any given statement in Column I can have correct matching with one or more statement(s) given in Column II . For example, if for a given question, statement B matches with thestatements given in q and r, then for that particular question, against statement B, darken the bubblescorresponding to q and r in the ORS.

37. Match the statements in Column-I with those in Column-II.[Note: Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginarypart and the real part of z.]

Column I Column II

(A) The set of points z satisfying |z i|z||= |z +i|z|| is contained in or equal to

(p) an ellipse with eccentricity 4/5

(B) The set of points z satisfying |z + 4| + |z 4| =10 is contained in or equal to

(q) the set of points z satisfying Im z = 0

(C)If |w| = 2, then the set of points z = w -

1w

is

contained in or equal to

(r) the set of points z satisfying |im z| 1

(D)If |w| = 1, then the set of points z = w + 1w is

contained in or equal to

(s)the set of points z satisfying |Re z|

2

(t) the set of points z satisfying |z| 3

Key. (A-q, r), (B-p), (C-p, s), (D-q, r, s, t)


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Sol. |z i|z|| = |z + i|z||

(A) Putting z = x + iy

We get y 2 2x y 0+ = i.e., Im (z) = 0.

(B) 2ae = 8, 2a = 10

410e 8 e 5= = (0, 3)

(0, 3)

(5, 0) (5, 0)(0, 0)

2 16b 25 1 925

= =

2 2x y1

25 9

+ =

(C) 1

z 2(cos isin )2(cos i sin )

= + +

12(cos isin ) (cos i sin )

2= +

3 5z cos i sin

2 2= +

( )( )3

, 02

-3

,02

Let z = x + iy

3 5x cos , y sin

2 2= =

2 22x 2y

13 5

+ =

2 2x y9 25 14 4

+ =

29 25

(1 e )4 4=

2 9 16e 125 25

= =

4

e5

=


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(D) Let w = cos + isin 1

z x iy ww

= + = +

x iy 2cos + = x 2cos , y 0= =

(q), (s)

38. Match the statements in Column I with the values in Column-II

Column I Column II

(A) A line from the origin meets the lines8

xx 2 y 1 z 1 y 3 z 13and1 2 1 2 1 1

+ + = = = =

at

P and Q respectively. If length PQ = d, thend2 is

(p) -4

(B) The values of x satisfying tan -1(x + 3) tan -1

(x 3) = sin -1 35

are

(q) 0

(C) Non-zero vectors a,b and c

satisfy a.b = 0,

(b a).(b c) 0 + =

and 2 | b c | | b a |+ =

. If

a b 4c,= +

then the possible values of are

(r) 4

(D) Let f be the function on [- , ] given by f(0) =

9 and f(x) = sin9x x

/sin2 2

for x 0. The

value of 2

f(x)dx

is

(s) 5

(t) 6

Key. (A t), (B - p, r), (C - q), (D - r)

Sol. (A) P( 2,1 2 , 1) +

8Q 2 , 3, 1

3 + +

22 2 2 22| PQ | d 2 ( 2 4) ( 2)

3 = = + + +

As OP and OQ

are collinear8 1 2 1

23 3 1

+ = = +

(from last two)

2 0 + = . . .(i)

and5 14

43 3

+ = (from Ist and IIIrd) . . .(ii)

from (i) and (ii) 2 3 5 = . . . (iii)


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from (i) and (iii) 3 2 + 2 1 = 0

1

1,3

=

so, = 1, 3

Hence, 2109

d or 6

9

=

(B) 1 1 13

tan (x 3) tan (x 3) sin5

+ =

1 1 1 3tan (x 3) tan (x 3) tan4

+ =

Let tan 1(x + 3) = , tan 1(x 3) = tan x 3, tan x 3 = + =

3tan( )

4 =

tan tan 3

1 tan tan 4

= =+

2

(x 3) (x 3) 31 x 9 4+ = =

+

2

6 3x 8 4

= =

2x 8 8= = 2x 16= =

x = 4

(C) (b a).(b c) 0 + =

Puta b

C 4

=

a b(b a). b 0

4

+ =

(b a).((4 )b a) 0 + =

2 2(4 )| b | | a | 0 =

a.b 0= 2 2(4 )| b | | a | = ( a.b 0)=

Alsoa b

2 | b c | | b a | again put C4

+ = =

a b2 b | b a |

4 + =

1| (4 )b a | | b a |

2 + =


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2 2 2 2 2(4 ) | b | | a | 4 | b | 4 | a | sin ce a.b 0 + = + =

( )2 2 2(4 ) 4 | b | 3 | a | =

( )2 2 2(4 ) 4 | b | 3(4 ) | b | = (4 )2 4 = 12 3 16 + 2 8 4 = 12 3 2 5 = 0 = 0 or 5. but = 5 is not satisfying so = 0.

(D)

9xsin

2 2x

sin2

0

9x x2sin cos

4 2 2x x

2sin cos2 2

=

0 0

1 2

4 sin5x 4 sin 4xsin x sin xI I

= +

I1 = , I2 = 0

So,4

4 =

.


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PART II I: PHYSI CS



39. A block of mass 2 kg is free to move along the xaxis. It is at rest and from t =0 onwards it is subjected to a timedependent force F(t) in the x direction. Theforce F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is(A) 4.50 J (B) 7.50 J(C) 5.06 J (D) 14.06 J.

F(t)

Ot

3s4.5s

4N

Key. (C)

Sol. 1 1

2(V 0) 4 3 2 1.52 2

=

2V 6 1.5= 4.5

V2

= 2

1 9K (2)2 4

=

815.06 J

16= = .


40. A uniformly charged thin spherical shell of radius R carries uniformsurface charge density of per unit area. It is made of two hemisphericalshells, held together by pressing them with force F (see figure). F isproportional to

(A) 2 2

0

1R

(B) 2

0

1R

F F

(C)2

01 R (D)

2

20

1 R .

Key. (A) Sol.

2

20

F2 R =

2 2

0

RF

2 =

.


41. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field

of strength 581

107 Vm 1. When the field is switched off, the drop is observed to fall with terminal

velocity 2 10 3 ms1. Given g = 9.8 ms 2, viscosity of the air = 1.8 10 5 Ns m 2 and the density of oil =900 kg m 3 , the magnitude of q is


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(A) 191.6 10 C (B) 193.2 10 C (C) 194.8 10 C (D) 198.0 10 C .

Key. (D) Sol. mg =f vise

3T

4R g 6 Rv

3 =

5Tv9 3R 10 m2 g 7

= =

Eq = mg34 R g

q3 E

= 19q 8 10 C= .

42. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale whichmatch with 16 main scale divisions. For this Vernier calipers, the least count is(A) 0.02 mm (B) 0.05 mm(C) 0.1 mm (D) 0.2 mm.

Key. (D) Sol. V.C. = 1 div of M.S. 1 div V.S.

161 div of M.S. div of M.S.20=

4div of M.S.

20=

0.2 mm= .Hence correct option is (D).

43. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and themirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is(A) virtual and at a distance of 16 cm from the mirror(B) real and at a distance of 16 cm from the mirror(C) virtual and at a distance of 20 cm from the mirror(D) real and at a distance of 20 cm from the mirror.

Key. (B) Sol. For lens1 1 1V 30 15

=

V = 30 cmFor mirror

u = + 20 cmv = 20 cm

For lensu = 101 1 1V 10 15

=

1 3 2

V 30

+=

V = 6 cm from lens = 16 cm from mirror.Hence correct option is (B).

44. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibratingin its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wireis 50 N and the speed of sound is 320 ms 1, the mass of the string is(A) 5 grams (B) 10 grams(C) 20 grams (D) 40 grams.


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Key. (B)

Sol. 320 2 50

4 0.8 2 0.5=

320 508 0.8

=

5050 50 =

1k g / m

50 =

1m 0.5 1000 gm

50=

10= grams.Hence correct option is (B).



45. A large glass slab ( = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It isseen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is thevalue of R ?

Key. 6. Sol.

R

8

3 3

sin tan5 4

= =

R 3 R 6 cm8 4

= = .

46. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is

observed to move from25 50

m to m3 7

in 30 seconds. What is the speed of the object in km per hour ?

Key. 3.

Sol. 1 1 1v u f

+ =

1

3 1 225 u 20

+ = +

1

1 10 12

u 100

+ =

1u 50 m=

2

7 1 250 u 20

+ =

2

1 10 14 1u 100 25

= =


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2u 25 m= 25 18 5 18

v 3 km / hr30 5 30

= = = .

47. To determine the half life of a radioactive element, a student

plots a graph of dN(t)

ndt

versus t. HeredN(t)

dtis the rate of

radioactive decay at time t. If the number of radioactivenuclei of this element decreases by a factor of p after 4.16years, the value of p is

1

2

34

5

6

2 3 4 5 6 7 8Years

n

D N ( t )

d t

Key. 8.

Sol. dN 1

n tdt 2

= 1

t2dN e

dt

=

11 year

2

=

12

0.69T 1.38 years

12

= =

4.16 years 3 half livesp = 8.

48. A diatomic ideal gas is compressed adiabatically to1

32of its initial volume. In the initial temperature of

the gas is T i (in Kelvin) and the final temperature is aT i, the value of a isKey. 4.

Sol. 1

1i i

1T v T v

32

=

7 2 2t 55 5 5(32) (32) (2 ) 4 = = = = .

49. At time t = 0, a battery of 10 V is connected across points A and Bin the given circuit. If the capacitors have no charge initially, at whattime (in seconds) does the voltage across them become 4 V ?[Take : n 5 1.6, n3 1.1= = ].

A B

2M 2 F

2 F2M Key. 4.

Sol. t / RC 602 2

v v (1 e ), R 102 2

= = = +

t / 4 64 10(1 e ) C 4 10 F = = t/ 4

e 1.6

= n5 n3 0.5 = = .




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Paragraph for Quest ion Nos . 50 to 52

When liquid medicine of density is to be put in the eye, it is done with the help of a dropper. As the bulbon the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate thesize of the drop. We first assume that the drop formed at the opening is spherical because that requires aminimum increase in its surface energy. To determine the size, we calculate the net vertical force due to thesurface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the

drop, the drop gets detached from the dropper.

50. If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r


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This section contains 2 questions . Each question four statements (A, B, C and D) given in Column I and fivestatements (p, q, r, s and t) in Column II . Any given statement in Column I can have correct matching with one or more statement(s) given in Column II . For example, if for a given question, statement B matches with thestatements given in q and r, then for that particular question, against statement B, darken the bubbles

corresponding to q and r in the ORS.

56. Two transparent media of refractive indices 1 and 3 have a solid lens shaped transparent material of refractive index 2 between them as shown in figures in Column II. A ray traversing these media is alsoshown in the figures. In Column I different relationship between 1, 2 and 3 are given. Match them to theray diagrams shown in Column II.

Column I Column II

(A) 1 2 < (p)

3 2 1

(B) 1 2 > (q)

3 2 1

(C) 2 3 = (r)

3 2 1

(D) 2 3 > (s)

3 2 1

(t)

3 2 1

Key. (A) (p), (r); (B) (q), (s), (t); (C) (p), (r), (t); (D) (q), (s)


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57. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltagesource (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) indifferent ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows throughthe circuit, the corresponding voltage V 1 and V 2. (indicated in circuits) are related as shown in Column I.Match the two :

Column I Column II

(A) I 0, V 1 is proportional to I (p) V 1 V2

6mH 3 F

V (B) I 0, V 2 > V 1 (q) V 1 V2

6mH 2

V (C) V 1 = 0, V 2 > V 1 (r) V 1 V2

6mH 2

V~ (D) I 0, V 2 is proportional to I (s) V 1 V2

6mH 3 F

V~ (t) V 1 V2

1k 3 F

V~ Key. (A) (r), (s), (t); (B) (q), (r), (s), (t); (C) (p), (q); (D) (r), (s), (t)

Sol. 3 1LX 2 (50)(6 10 ) 6 10 = =

4

C 6

1 106 10X

3 100 32 50 3 10 = = =

C LX X> .

Narayana Solutions iit jee 2010

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